# 10.8: Study Points

• If two opposing chemical reactions proceed simultaneously at the same rate, the processes are said to be in equilibrium. The two opposing reactions are shown linked with a double arrow (⇄). An example of opposing chemical reactions and their equilibrium expressions are:
N2(g) + 3 H2(g) → 2 NH3(g) and 2 NH3(g) → N2(g) + 3 H2(g)

The equilibrium would be written as: N2(g) + 3 H2(g) ⇄ 2 NH3(g)

• An equilibrium constant (K) is a numerical value that relates the concentrations of the products and reactants for a chemical reaction that is at equilibrium. The numeric value of an equilibrium constant is independent of the initial concentrations of reactants, but is dependent on the temperature. Equilibrium constants are generally not written with units (although they may be).
• Because equilibrium constants are written as the concentrations (or partial pressures) of products divided by the concentrations (or partial pressures) of reactants, a large value of K means that there are more products in the equilibrium mixture than there are products. Likewise, a small value of K means that, at equilibrium, there are more reactants than products.
• Equilibrium constants that are based on partial pressures are often written as KP, while equilibrium constants based on molar concentrations are written as Kc.
• An expression for an equilibrium constant can be written from a balanced chemical equation for the reaction. The Law of Mass Action states the following regarding equilibrium expressions:
• Partial pressures (or molar concentrations) of products are written in the numerator of the expression and the partial pressures (or concentrations) of the reactants are written in the denominator.
• If there is more that one reactant or more that one product, the partial pressures (or concentrations) are multiplied together.
• The partial pressure (or concentration) of each reactant or product is then raised to the power that numerically equals the stoichiometric coefficient appearing with that term in the balanced chemical equation.
• Reactants or products that are present as solids or liquids do not appear in the equilibrium expression.
• As an example of an equilibrium expression, consider the reaction of nitrogen and hydrogen to form ammonia. The partial pressure of ammonia will be in the numerator, and it will be squared. Because there are two reactants, the partial pressures for nitrogen and hydrogen will be multiplied in the denominator. The partial pressure of nitrogen will be raised to the “first power” (which is not shown) and the partial pressure of hydrogen will be cubed.

N2 (g) + 3 H2 (g) ⇄ 2 NH3 (g)

$\frac{(P_{NH_{3}})^{2}}{P_{N_{2}}(P_{H_{2}})^{3}}=K$

• If equilibrium values for a given reaction are known, the equilibrium constant can be calculated simply by substituting those values in the equilibrium expression. Quite often, however, initial and equilibrium values are only given for selected reactants and products. In these cases, initial and equilibrium values are arranged in an ICE Table, and the changes between initial and equilibrium states are calculated based on reaction stoichiometry.
• Because the partial pressure of a gas and the molar concentration of that gas are directly proportional, the ideal gas law can be rearranged as follows, to give an expression relating molarity and partial pressure of a gas.

$P_{gas}V=nRT$

Dividing by the volume in liters gives the term ${\displaystyle \left({}^{n}\!\!\diagup \!\!{}_{V_{liters}}\;\right)}$ which is equivalent to molarity, M.

$P_{gas}=MRT$

• Le Chatelier's Principle states that, if a "stress" is applied to a chemical reaction at equilibrium, the system will readjust in the direction that best reduces the stress imposed on the system. In this context, stress refers to a change in concentration, a change in pressure or a change in temperature, although only concentration is considered here. If pressure or temperature are changed, the numeric value K will change; if only concentration changes are involved, K does not change. In a reaction at equilibrium, the introduction of more products will shift the mass balance towards more reactants, and the introduction of more reactants will lead to the formation of more products, but the ratio of Products/Reactants (as defined by the equilibrium expression) does not change, hence, K is unchanged.
• For a weak acid, dissociating in water to form it's conjugate base and hydronium ion, the equilibrium constant is referred to as Ka. Because a weak acid is only "partially dissociated", the concentration of BH in solution is large, thus Ka for a weak acid is "small" (in the range of 10-3 to 10-6). For example, acetic acid (the acidic component of vinegar), has an acid dissociation constant of Ka = 1.8 x 10-5.

CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+(aq)

• For a solution of a weak acid in water, the concentration of hydronium ion will be very small. If the concentration of the weak acid is fairly large (typically > 0.01 M) the concentration of the undissociated acid will be much larger than [H3O+]. Because of this, the hydronium ion concentration (hence, the pH) can be fairly accurately estimated from the Ka of the weak acid and the initial concentration of the acid (Co), by the equation:

$[H_{3}O^{+}]=\sqrt{K_{a}\times C_{0}}$

• The equilibrium constant defining the solubility of an ionic compound with low solubility is defined as Ksp, where “sp” refers to “solubility product”. Because reactants or products that are present as solids or liquids do not appear in equilibrium expressions, for silver chloride, the expression Ksp will be written as:

AgCl(s) ⇄ Ag+(aq) + Cl-(aq)

$K_{sp}=[Ag^{+}][Cl^{-}]$

• For silver chloride, the solubility at 25 oC is 1.67 × 10-5 M. That means the concentrations of silver and chloride ions in solution are each 1.67 × 10-5 M, making the value of Ksp under these conditions: $K_{sp}=[Ag^{+}][Cl^{-}]=(1.67\times 10^{-5})^{2}=2.79\times 10^{-10}$

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